13) ∫₁⁶ 1/√(3x - 2) dx

Explanation:

Step1: Use substitution

Let $u = 3x - 2$, then $du=3dx$, and $dx=\frac{1}{3}du$. When $x = 1$, $u=3\times1 - 2=1$; when $x = 6$, $u=3\times6 - 2 = 16$.

Step2: Rewrite the integral

The integral $\int_{1}^{6}\frac{1}{\sqrt{3x - 2}}dx$ becomes $\frac{1}{3}\int_{1}^{16}u^{-\frac{1}{2}}du$.

Step3: Integrate using power - rule

The antiderivative of $u^{-\frac{1}{2}}$ is $2u^{\frac{1}{2}}$ (since $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$, $n\neq - 1$, here $n=-\frac{1}{2}$). So $\frac{1}{3}\int_{1}^{16}u^{-\frac{1}{2}}du=\frac{1}{3}\times2u^{\frac{1}{2}}\big|_{1}^{16}$.

Step4: Evaluate the definite integral

$\frac{2}{3}( \sqrt{16}-\sqrt{1})=\frac{2}{3}(4 - 1)$.

Step5: Calculate the result

$\frac{2}{3}\times3 = 2$.

Answer:

$2$